Normalmente, este problema se resuelve con el uso del módulo de un número en un bucle o convierte un número en una cadena. Para convertir un número en una cadena, puede usar la función itoa, por lo que considera la variante con el módulo de un número en un bucle.
contenido de un archivo get_digits.c
$ cat get_digits.c
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
// return a length of integer
unsigned long int get_number_count_digits(long int number);
// get digits from an integer number into an array
int number_get_digits(long int number, int **digits, unsigned int *len);
// for demo features
void demo_number_get_digits(long int number);
int
main()
{
demo_number_get_digits(-9999999999999);
demo_number_get_digits(-10000000000);
demo_number_get_digits(-1000);
demo_number_get_digits(-9);
demo_number_get_digits(0);
demo_number_get_digits(9);
demo_number_get_digits(1000);
demo_number_get_digits(10000000000);
demo_number_get_digits(9999999999999);
return EXIT_SUCCESS;
}
unsigned long int
get_number_count_digits(long int number)
{
if (number < 0)
number = llabs(number);
else if (number == 0)
return 1;
if (number < 999999999999997)
return floor(log10(number)) + 1;
unsigned long int count = 0;
while (number > 0) {
++count;
number /= 10;
}
return count;
}
int
number_get_digits(long int number, int **digits, unsigned int *len)
{
number = labs(number);
// termination count digits and size of a array as well as
*len = get_number_count_digits(number);
*digits = realloc(*digits, *len * sizeof(int));
// fill up the array
unsigned int index = 0;
while (number > 0) {
(*digits)[index] = (int)(number % 10);
number /= 10;
++index;
}
// reverse the array
unsigned long int i = 0, half_len = (*len/2);
int swap;
while (i < half_len) {
swap = (*digits)[i];
(*digits)[i] = (*digits)[*len - i - 1];
(*digits)[*len - i - 1] = swap;
++i;
}
return 0;
}
void
demo_number_get_digits(long int number)
{
int *digits;
unsigned int len;
digits = malloc(sizeof(int));
number_get_digits(number, &digits, &len);
printf("%ld --> [", number);
for (unsigned int i = 0; i < len; ++i) {
if (i == len - 1)
printf("%d", digits[i]);
else
printf("%d, ", digits[i]);
}
printf("]\n");
free(digits);
}
demo con la GNU GCC
$~/Downloads/temp$ cc -Wall -Wextra -std=c11 -o run get_digits.c -lm
$~/Downloads/temp$ ./run
-9999999999999 --> [9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9]
-10000000000 --> [1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
-1000 --> [1, 0, 0, 0]
-9 --> [9]
0 --> [0]
9 --> [9]
1000 --> [1, 0, 0, 0]
10000000000 --> [1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
9999999999999 --> [9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9]
demo con el LLVM/Clang
$~/Downloads/temp$ rm run
$~/Downloads/temp$ clang -std=c11 -Wall -Wextra get_digits.c -o run -lm
setivolkylany$~/Downloads/temp$ ./run
-9999999999999 --> [9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9]
-10000000000 --> [1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
-1000 --> [1, 0, 0, 0]
-9 --> [9]
0 --> [0]
9 --> [9]
1000 --> [1, 0, 0, 0]
10000000000 --> [1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
9999999999999 --> [9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9]
entorno de pruebas
$~/Downloads/temp$ cc --version | head -n 1
cc (Debian 4.9.2-10) 4.9.2
$~/Downloads/temp$ clang --version
Debian clang version 3.5.0-10 (tags/RELEASE_350/final) (based on LLVM 3.5.0)
Target: x86_64-pc-linux-gnu
Thread model: posix
Dado que no he encontrado en mi compendio operador de poco personal (http://graphics.stanford.edu/~seander/bithacks.html), No creo que esto sea posible sin una elaboración más profunda. – phimuemue
si son decimales, no puede obtenerlos usando bitwise. si son hexadecimales, entonces es posible. Por favor especifica. – Andrey